Problem 89 - Chemistry The Central Science

Question

At \(700 \mathrm{K},\) the equilibrium constant for the reaction $$\operatorname{CCI}_{4}(g) \Longrightarrow \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g)$$ is \(K_{p}=0.76 .\) A flask is charged with 2.00 atm of \(\mathrm{CCl}_{4}\) ,which then reaches equilibrium at 700 \(\mathrm{K}\) . (a) What fraction of the CCl\(_{4}\) is converted into \(\mathrm{C}\) and \(\mathrm{Cl}_{2} ?(\mathbf{b})\) what are the partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) at equilibrium?

Step-by-Step Solution

Answer

At 700 K, the equilibrium constant for the reaction is \(K_p = 0.76\). When a flask containing 2.00 atm of CCl\(_4\) reaches equilibrium, 50.7% of the CCl\(_4\) is converted into C and Cl\(_2\). At equilibrium, the partial pressures are 1.493 atm for CCl\(_4\) and 1.014 atm for Cl\(_2\).

Step 1: Write down the balanced chemical equation

The given reaction is: \[CCl_4(g) ⇄ C(s) + 2Cl_2(g)\]

Step 2: Express the equilibrium constant (Kp) in terms of partial pressures

For the given reaction, the equilibrium constant expression is: \[K_p = \frac{P_{Cl_2}^2}{P_{CCl_4}}\]

Step 3: Set up an equilibrium pressure table (ICE table)

Let x represent the fraction of CCl\(_4\) that dissociates. We set up the following table: | | Initial | Change | Equilibrium | |---------|---------|---------|-----------------| | CCl\(_4\) | 2.00 atm| -x | 2.00 - x | | Cl\(_2\) | 0 atm | 2x | 2x |

Step 4: Substitute equilibrium pressures in Kp expression

Using the information from the table, we can write the Kp expression as: \[K_p = \frac{(2x)^2}{2.00 - x}\] We are given that \(K_p = 0.76\). So, we have: \[0.76 = \frac{(2x)^2}{2.00 - x}\]

Step 5: Solve for x (fraction of CCl4 converted)

We can now solve this equation for x: \[0.76 (2.00 - x) = (2x)^2\] \[1.52 - 0.76x = 4x^2\] Rearrange the equation: \[4x^2 + 0.76x - 1.52 = 0\] Now solve the quadratic equation for x either by factoring, completing the square, or using the quadratic formula. Using the quadratic formula, we find: \[x = 0.507 (rounded)\] This means 50.7% of the CCl\(_4\) is converted to C and Cl\(_2\) at equilibrium.

Step 6: Calculate the partial pressures at equilibrium

Now we can find the equilibrium partial pressures using the equilibrium pressure values from the table: Partial pressure of CCl\(_4\): \[P_{CCl_4} = 2.00 - x = 2.00 - 0.507 = 1.493 \, atm\] Partial pressure of Cl\(_2\): \[P_{Cl_2} = 2x = 2(0.507) = 1.014 \, atm\] Therefore, at equilibrium, the partial pressures are 1.493 atm for CCl\(_4\) and 1.014 atm for Cl\(_2\).

Key Concepts

Partial Pressure

In a chemical reaction involving gases, partial pressure is the pressure that each gas in a mixture would exert if it occupied the entire volume alone. It's crucial in studying reactions like the decomposition of carbon tetrachloride (\(CCl_4\)). In the exercise, different partial pressures are considered to determine the equilibrium state of the system.
When dealing with equilibrium, each gas's partial pressure contributes to the total pressure used in the equilibrium constant expression. In the given chemical reaction, the partial pressure of \(CCl_4\) decreases as it decomposes, while the partial pressure of \(Cl_2\) increases since it's a product.
Here’s why these values matter:

• The equilibrium constant \(K_p\) depends on the partial pressures of the gases involved.
• The pressures help predict how much of the reactant converts to the product at equilibrium.

Understanding partial pressure allows you to relate changes in gas concentrations with shifts in equilibrium, vital for predicting the outcomes of reactions.

Chemical Reaction

A chemical reaction can be seen as a process where reactants transform into products. In the exercise, the reaction involves the decomposition of \(CCl_4\) into \(C\) and \(Cl_2\). Notice that \(C\) is a solid and does not appear in the equilibrium expression, as only gaseous substances are considered when dealing with \(K_p\).
Each reaction at equilibrium has an associated equilibrium constant, which shows the ratio of the concentration (or partial pressures) of products to reactants. For our reaction:

• The formula for \(K_p\) is based on the balanced chemical equation.
• It includes only the gaseous components, hence the absence of \(C(s)\).
• The equilibrium constant helps to understand the extent to which a reaction proceeds.

Knowing the balanced equation allows us to set up the calculation framework necessary for determining how much reactant is converted into products, which directly ties into calculating reactant and product quantities at equilibrium.

ICE Table

An ICE table is a useful tool for understanding how chemical concentrations change as a reaction progresses towards equilibrium. ICE stands for Initial, Change, and Equilibrium.
Let's break it down:

• Initial: Start with the known initial pressures or concentrations of reactants and products.
• Change: Represents the change as the reaction reaches equilibrium. This is usually expressed in terms of a variable, often \(x\), representing the amount the reactants and products change by.
• Equilibrium: Shows the pressures or concentrations at equilibrium, which are calculated based on the changes identified.

In the exercise, the ICE table sets the scene for solving the equilibrium problem by lining out where each component stands at the start, the shifts they undergo, and their positions in the end.
Mastering the ICE table approach is essential for handling equilibrium problems effectively, as it allows for a systematic method to predict equilibrium conditions from initial states.